InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
\(\frac{{sin\theta [\left( {1 - tan\theta } \right)tan\theta + {{\sec }^2}\theta ]}}{{\left( {1 - \sin \theta } \right)\tan \theta \left( {1 + tan\theta } \right)\left( {sec\theta + tan\theta } \right)}}\) is equal to:1. sinθ cosθ 2. -13. cosecθ secθ 4. 1 |
|
Answer» Correct Answer - Option 4 : 1 GIVEN: \(\frac{{sinθ [\left( {1 - tanθ } \right)tanθ + {{\sec }^2}θ ]}}{{\left( {1 - \sin θ } \right)\tan θ \left( {1 + tanθ } \right)\left( {secθ + tanθ } \right)}}\) FORMULA USED: \({\sec ^2}θ - \;{\tan ^2}θ = 1\), sin2θ + cos2θ = 1. tanθ = sinθ/cosθ, secθ = 1/cosθ CALCULATION: \(\frac{{sinθ [\left( {1 - tanθ } \right)tanθ + {{\sec }^2}θ ]}}{{\left( {1 - \sin θ } \right)\tan θ \left( {1 + tanθ } \right)\left( {secθ + tanθ } \right)}}\) \( ⇒ \;\frac{{\sin θ \;[\tan θ - \;{{\tan }^2}θ + \;1 + \;{{\tan }^2}θ ]}}{{(1 - \;\sin θ )\;\tan θ \;(1 + \;\tan θ )\;(\sec θ + \;\tan θ )}}\) \( ⇒ \frac{{\sin θ \;(1 + \;\tan θ )}}{{(1 - \sin θ )\;\tan θ \;(1 + \;\tan θ )\;(\sec θ + \;\tan θ )}}\) \(⇒ \;\frac{{\sin θ }}{{(1 - \;\sin θ )\;\frac{{\sin θ }}{{\cos θ }}\;(\sec θ + \;\tan θ )}}\) \(⇒ \;\frac{{\cos θ }}{{(1 - \;\sin θ )\;\left( {\frac{1}{{\cos θ }} + \;\frac{{\sin θ }}{{\cos θ }}} \right)}}\) \( ⇒ \frac{{{{\cos }^2}θ }}{{(1 - \sin θ )\;(1 + \;\sin θ )}}\) \(⇒ \frac{{{{\cos }^2}θ }}{{(1 - \;{{\sin }^2}θ )}}\) \(⇒ \;\frac{{{{\cos }^2}θ }}{{{{\cos }^2}θ }}\) ⇒ 1 |
|